Cartesian Product of Intervals is Convex Set

Theorem

Let $n \in \N$.

For all $k \in \set {1, \ldots, n}$, let $\Bbb I_k$ be a real interval of any of the real interval types.

Then the cartesian product $\Bbb I_1 \times \ldots \times \Bbb I_n$ is a convex set.

Let $\mathbf x, \mathbf y \in \Bbb I_1 \times \ldots \times \Bbb I_n$ with:

$\ds \mathbf x$

$=$

$\ds \tuple {x_1, \ldots, x_n}$

$\ds \mathbf y$

$=$

$\ds \tuple {y_1, \ldots, y_n}$

where $x_k , y_k \in \Bbb I_k$ for all $k \in \set {1, \ldots, n}$.

Let $t \in \closedint 0 1$.

Suppose that $x_k \le y_k$.

It follows that:

$\ds x_k$

$=$

$\ds t x_k + \paren {1-t} x_k$

$\ds $

$\le$

$\ds t x_k + \paren {1-t} y_k$

as $1-t \in \closedint 0 1$

$\ds $

$\le$

$\ds t y_k + \paren {1-t} y_k$

$\ds $

$=$

$\ds y_k$

By definition of real interval, it follows that $t x_k + \paren {1-t} y_k \in \Bbb I_k$.

If instead $x_k \ge y_k$, it follows by similar calculations that:

$x_k \ge t x_k + \paren {1-t} y_k \ge y_k$

By definition of real interval, it follows that $t x_k + \paren {1-t} y_k \in \Bbb I_k$.

It follows that $t \mathbf x + \paren {1-t} \mathbf y \in \Bbb I_1 \times \ldots \times \Bbb I_n$.

By definition of convex set, it follows that $\Bbb I_1 \times \ldots \times \Bbb I_n$ is a convex set.

$\blacksquare$